Question: Simplify and expand the following expression: $ \dfrac{5}{r - 6}-\dfrac{2r}{3r - 2} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(r - 6)(3r - 2)$ Multiply the first term by $\dfrac{3r - 2}{3r - 2}$ $ \begin{align*} \dfrac{5}{r - 6} \times \dfrac{3r - 2}{3r - 2} & = \dfrac{(5)(3r - 2)}{(r - 6)(3r - 2)} \\ & = \dfrac{15r - 10}{(r - 6)(3r - 2)}\end{align*} $ Multiply the second term by $\dfrac{r - 6}{r - 6}$ $ \begin{align*} \dfrac{2r}{3r - 2} \times \dfrac{r - 6}{r - 6} & = \dfrac{(2r)(r - 6)}{(3r - 2)(r - 6)} \\ & = \dfrac{2r^2 - 12r}{(3r - 2)(r - 6)}\end{align*} $ Now we have: $ = \dfrac{15r - 10}{(r - 6)(3r - 2)} - \dfrac{2r^2 - 12r}{(3r - 2)(r - 6)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{15r - 10 - (2r^2 - 12r)}{(r - 6)(3r - 2)} $ $ = \dfrac{15r - 10 - 2r^2 + 12r}{(r - 6)(3r - 2)} $ $ = \dfrac{27r - 10 - 2r^2}{(r - 6)(3r - 2)}$ Expand the denominator: $ = \dfrac{27r - 10 - 2r^2}{3r^2 - 20r + 12}$